Answer: B (Longitudinal waves travel through solids, liquids, and gases, Transverse waves only travel through solids)
Explanation: A longitudinal wave alternately compresses the medium and stretches it out. Solids, liquids and gasses all push back when they are compressed— so they are all able to store energy this way and thus transmit the wave. But, Transverse waves can only go through solids because they have enough shear strength, but liquids and gases don't. ( Transverse waves are the transfer of energy in a motion that is perpendicular to the direction the wave is traveling. Only solids are able to switch it's motion to travel through the wave.)
Answer:
b
Explanation:
1. a) If a particle's position is given by LaTeX: x\:=\:4-12t\:+\:3t^2x = 4 − 12 t + 3 t 2(where t is in seconds and x is in meters), what is its velocity at LaTeX: t=1st = 1 s? b) Is it moving in the positive or negative direction of LaTeX: xx just then? c) What is its speed just then? d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculations.) e) Is there ever an instant when the velocity is zero? If so, give the time LaTeX: tt ; if not, answer no. f) Is there a time after LaTeX: t=3st = 3 s when the particle is moving in the negative direction of LaTeX: xx? If so, give the time LaTeX: tt; if not, answer no. (Hint: Speed= LaTeX: \mid v\mid∣ v ∣)
Answer:
a) v=-6m/s
b) negative direction
c) 6m/s
d) decreasing
e) for t=2s
f) Yes
Explanation:
The particle position is given by:
[tex]x=4-12t+3t^2[/tex]
a) the velocity of the particle is given by the derivative of x in time:
[tex]v=\frac{dx}{dt}=-12+6t[/tex]
and for t=1s you have:
[tex]v=\frac{dx}{dt}=-12+6(1)^2=-6\frac{m}{s}[/tex]
b) for t=1s you can notice that the particle is moving in the negative x direction.
c) The speed can be computed by using the formula:
[tex]|v|=\sqrt{(-12+6t)^2}=\sqrt{(-12+6)^2}=6\frac{m}{s}[/tex]
d) Due to the negative value of the velocity in a) you can conclude that the speed is decreasing.
e) There is a time in which the velocity is zero. You can conclude that because if t=2 in the formula for v in a), v=0
[tex]v=0=-12+6(t)\\\\t=\frac{12}{6}=2[/tex]
f) after t=3s the particle will move in the negative direction, this because it is clear that 4+3t^2 does not exceed -12t.
Answer:
a) v (1) = -6 m/s
b) negative x-direction
c) s ( 1 ) = 6 m/s
d) The speed decreases at t increases from 0 to 2 seconds.
e) At t = 2 s, the velocity is 0
f) No
Explanation:
Given:-
- The position function of the particle:
x (t) = 4 - 12t + 3t^2
Find:-
what is its velocity at t = 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer no. (f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer no.
Solution:-
- The velocity function of the particle v(t) can be determined from the following definition:
v (t) = d x(t) / dt
v (t) = -12 + 6t
- Evaluate the velocity at time t = 1 s:
v (1) = -12 + 6(1)
v (1) = -6 m/s
- The negative sign of the velocity at time t = 1s shows that the particle is moving in the negative x-direction.
- The speed ( s ( t )is the absolute value of velocity at time t = 1s:
s ( t ) = abs ( v ( t ) )
s ( 1 ) = abs ( v ( 1 ) )
s ( 1 ) = abs ( -6 )
s ( 1 ) = 6 m/s
- The speed of the particle at time t = 0,
s ( t ) = abs ( -12 + 6t )
s ( 0 ) = abs (-12 + 6 (0) )
s ( 0 ) = abs ( -12 )
s ( 0 ) = 12 m/s
- The speed of the particle at time t = 2,
s ( t ) = abs ( -12 + 6t )
s ( 2 ) = abs (-12 + 6 (2) )
s ( 2 ) = abs ( 0 )
s ( 2 ) = 0 m/s
- Hence, the speed of the particle decreases from s ( 0 ) = 12 m/s to s ( 2 ) = 0 m/s in the time interval t = 0 to t = 2 s.
- As the speed decreases as time increases over the interval t = 0 , t = 2 s the velocity v(t) also approaches 0, at time t = 2 s. s ( 2 ) = 0 m/s.
- We will develop an inequality when v (t) is positive:
v (t) = -12 + 6t > 0
6t > 12
t > 2
- So for all values of t > 2 the velocity of the particle is always positive.
When light travels from a material with a lower index of refraction to a material with a higher index of refraction, the refracted beam will: a) shift away from the normal b) shift towards the normal
Answer:
Shift towards the normal
Explanation:
Refraction is defined as the change in direction of light rays when passing through from a medium to another.
The ray can either pass through a less dense medium to a denser medium or from a denser medium to a less dense medium. The light ray bends towards or away from the normal ray(ray perpendicular to the plane) depending whether the travels from less dense to denser or otherwise.
Note that if a ray travels from less dense medium which have a low refractive index like air to a more dense medium like water which have a higher refractive index than air, the refracted ray tends to bend towards the normal, otherwise they bend away from the normal.
Answer:
Shift towards the normal
Explanation:
In a summer storm, the wind is blowing with a velocity of 24 m/s north. Suddenly in 3 seconds the wind's velocity is 6 m/s. What is the wind's deceleration?
Answer:
The deceleration is [tex]6ms^{-2}[/tex]
Explanation:
Acceleration is change in velocity with respect to time.
[tex]a = \frac{\Delta V}{\Delta t}\\a = \frac{24-6}{3} \\a = 6[/tex]